Question: An equilateral triangle $ABC$ shares a common side $BC$ with a square $BCDE,$ as pictured. What is the number of degrees in $\angle DAE$ (not pictured)? [asy]
pair pA, pB, pC, pD, pE;
pA = (0, 0);
pB = pA + dir(300);
pC = pA + dir(240);
pD = pC + dir(270);
pE = pB + dir(270);
draw(pA--pB--pC--pA);
draw(pB--pC--pD--pE--pB);
label("$A$", pA, N);
label("$B$", pB, E);
label("$C$", pC, W);
label("$D$", pD, SW);
label("$E$", pE, SE);
[/asy]
Answer: First of all, for our benefit, we should draw in the desired angle: [asy]
pair pA, pB, pC, pD, pE;
pA = (0, 0);
pB = pA + dir(300);
pC = pA + dir(240);
pD = pC + dir(270);
pE = pB + dir(270);
draw(pA--pB--pC--pA);
draw(pB--pC--pD--pE--pB);
draw(pD--pA--pE, red);
label("$A$", pA, N);
label("$B$", pB, E);
label("$C$", pC, W);
label("$D$", pD, SW);
label("$E$", pE, SE);
[/asy] We can see that $AB = BC = EB,$ thus $\triangle ABE$ is an isosceles triangle, where $\angle ABE = 90^\circ + 60^\circ = 150^\circ.$ Since the other two angles are equal and add up to $30^\circ$, we have that $\angle BAE = 15^\circ.$ Likewise, $\angle CAD = 15^\circ.$

Then, $\angle DAE = \angle CAB - \angle CAD - \angle BAE = 60^\circ - 15^\circ - 15^\circ = \boxed{30^\circ.}$